Let $E, F \subseteq X$ such that $E \subseteq F$. If $f(E)$ is empty, then result R7: Empty set is subset of every set guarantees that $f(E) = \emptyset \subseteq f(E)$. Suppose thus that $f(E)$ is nonempty and fix $y \in f(E)$. Then $y = f(x)$ for some $x \in E$. Since $E \subseteq F$, then also $x \in F$ and thus $y = f(x) \in f(F)$. Since $y \in f(E)$ was arbitrary, then $y \in f(F)$ for every $y \in f(E)$ and hence, $f(E) \subseteq f(F)$. $\square$
